Question 579150
{{{f(x)= x^3-12x^2-55x+150}}} is a polynomial
We know that if it has a rational zero, it will be a "fraction" whose denominator is a factor of the leading coefficient and whose numerator is a factor of the constant (or independent term, whatever you call it), with a plus or minus sign in front.
The leading coefficient (the coefficient of the term of highest degree) is that invisible 1 in front of {{{x^3}}}. So the denominator is a factor of 1, which could only be 1. Any rational zero will be an integer. It has to be a factor of 150. There are 12 positive factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30,7 5, and 150. The same numbers with a minus sign in front are also possibilities. I tried 1, and it was not a zero, but then I tried 2, and 2 was a zero.
Since 2 is a zero of {{{f(x)}}}, {{{(x-2)}}} is a factor of {{{f(x)}}}. {{{f(x)}}} can be evenly divided by {{{(x-2)}}}.
I did the division and the result was {{{x^2-10-75}}}. So
{{{f(x)= x^3-12x^2-55x+150=(x-2)(x^2-10-75)}}}
Factoring {{{x^2-10-75}}}, I found that {{{x^2-10-75=(x-15)(x+10)}}}
So {{{f(x)= (x-2)(x-15)(x+10)}}} and its zeros are x=2, x=15, and x=-10.